# Accelerated frame approximation in Schwarzschild metric far from the horizon

It is clear to me that if I take the Schwarzschild metric $$ds^2 = \left(1-\frac{2M}{r}\right)dt^2 - \left(1-\frac{2M}{r}\right)^{-1} dr^2$$ and choose $$\rho = 2\sqrt{\frac{r}{2M} -1}$$ then I get the Rindler coordinates when $$\rho << 1$$.$$ds^2 = \frac{\rho^2}{4} dt^2 - 4M^2 d\rho^2$$ Therefore I can say that when I am close to the horizon of a black hole at constant $$\rho$$ (or $$r$$ equivalently) then I am constantly accelerating outwards and do my space-time diagrams in Rindler coordinates. But this approximation is not valid when $$\rho >> 1$$, i.e. when the field is weak.

In the weak field approximation we have $$ds^2 = \left(1-\frac{2M}{r}\right)dt^2 - \left(1+\frac{2M}{r}\right) dr^2$$ my intuition tells me that to stay at $$r$$ constant I have to be constantly accelerating outwards, but I can't find any coordinate transform that brings me to the Rindler coordinates from the weak field approximation. Please help :)

• It does not work in the exact metric, only if you take $r$ close to $2M$ (or equivalently $\rho << 1$) you end up with the Rindler metric at first order. – albe_rola Jan 27 '19 at 23:37

Before looking for a solution, we should think carefully about what kind of result we should expect. A metric of the form $$\rho^2 dt^2- d\rho^2 \tag{1}$$ represents flat spacetime, and any worldline with constant $$\rho$$ is undergoing constant acceleration (constant weight). Because it is flat, a metric of this form can't be a good approximation to the Schwarzschild metric for all $$\rho$$. We can only expect it to be a good approximation for $$\rho=\rho_0+\epsilon$$ for some given $$\rho_0$$ and with sufficiently small $$\epsilon$$. Near the horizon, we have such an approximation with $$\rho_0=0$$; the hovering observer's acceleration diverges at the horizon, as it does in flat spacetime at the Rindler horizon. But away from the horizon, we will have $$\rho_0\neq 0$$, and then we can only expect a flat-spacetime approximation (1) to be valid modulo terms of order $$\epsilon^2$$.
With that in mind, start with the Schwarzschild metric in coordinates $$t$$ and $$\mathbf{x}\equiv (x,y,z)$$: $$d\tau^2=A(r)dt^2-\frac{dr^2}{A(r)}-(d\mathbf{x}^2-dr^2) \\ r\equiv\sqrt{x^2+y^2+z^2} \hskip2cm A(r)\equiv 1-\frac{2M}{r}. \tag{3}$$ Let $$\mathbf{x}=(0,0,z_0)$$ be the worldline of the hovering observer, and expand everything to first order in $$x$$, $$y$$, and $$\delta z\equiv z-z_0$$. This gives $$r^2\approx z_0^2+2z_0\delta z \hskip1cm \Rightarrow \hskip1cm r\approx z_0+\delta z=z \tag{4}$$ and $$A(r)\approx 1-\frac{2M}{z}\approx A(r_0)+\frac{2M}{z_0^2}\delta z \equiv a+bz. \tag{5}$$ All approximations are understood to be valid modulo terms of order $$(\delta z)^2$$. Use these in (3) to get $$d\tau^2\approx (a+bz)dt^2-\frac{dz^2}{a+bz}-dx^2-dy^2. \tag{6}$$ Now define $$\rho$$ by $$\rho^2\equiv a+bz \tag{7}$$ to get the Rindler-like form $$d\tau^2\approx \rho^2 dt^2-\frac{d\rho^2}{(b/2)^2}-dx^2-dy^2, \tag{8}$$ with the understanding that this is only valid in a neighborhood of $$\mathbf{x}= (0,0,z_0)$$, which corresponds to $$\rho_0=\sqrt{a+bz_0}$$. The approximation is good modulo terms of order $$\epsilon^2$$, where $$\epsilon\equiv\rho-\rho_0$$. This is the desired result.